sum from n=2 to infinity of-(2/3)^n

Lösung

n = 2 \sum \infty - (\frac{2}{3})^{n}

- \frac{4}{3}

Dezimale

- 1.33333 \dots

Schritte zur Lösung

n = 2 \sum \infty - (\frac{2}{3})^{n}

Wende die Regel der konstanten Multiplikation:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= - n = 2 \sum \infty (\frac{2}{3})^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= - (n = 0 \sum \infty (\frac{2}{3})^{n} - (\frac{2}{3})^{0} - (\frac{2}{3})^{1})

n = 0 \sum \infty (\frac{2}{3})^{n} = 3

= - (3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1})

Vereinfache

- (3 - (\frac{2}{3})^{0} - (\frac{2}{3})^{1}) : - \frac{4}{3}

= - \frac{4}{3}

x \to 0 + lim (\frac{ln ( 1 + 2 x )}{x})

\frac{d}{d x} (\frac{x + y}{2 + cos ( x )})

l a pl a ce t r an s f or m 2 t^{2} - e^{- t}

x \to \frac{π}{2} lim (cos^{2} (x))

f (x) = \frac{cos ( x ^{3} )}{e ^{x}}